若-3∈{a-3,2a-1,a²-4},求实数a的值.当a-3=-3时a=0当a=0时,2a-1=-1当a=0时,a²-4=-4此时集合为{-3,-1,-4},符合当2a-1=-3时,a=-1当a=-1时,a-3=-4当a=-1时,a²-4=-3此时集合为{-4,-3,-3}不满足集合中元素的互异
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![若-3∈{a-3,2a-1,a²-4},求实数a的值.当a-3=-3时a=0当a=0时,2a-1=-1当a=0时,a²-4=-4此时集合为{-3,-1,-4},符合当2a-1=-3时,a=-1当a=-1时,a-3=-4当a=-1时,a²-4=-3此时集合为{-4,-3,-3}不满足集合中元素的互异](/uploads/image/z/3818623-31-3.jpg?t=%E8%8B%A5-3%E2%88%88%7Ba-3%2C2a-1%2Ca%26%23178%3B-4%7D%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%80%BC.%E5%BD%93a-3%3D-3%E6%97%B6a%3D0%E5%BD%93a%3D0%E6%97%B6%2C2a-1%3D-1%E5%BD%93a%3D0%E6%97%B6%2Ca%26%23178%3B-4%3D-4%E6%AD%A4%E6%97%B6%E9%9B%86%E5%90%88%E4%B8%BA%7B-3%2C-1%2C-4%7D%2C%E7%AC%A6%E5%90%88%E5%BD%932a-1%3D-3%E6%97%B6%2Ca%3D-1%E5%BD%93a%3D-1%E6%97%B6%2Ca-3%3D-4%E5%BD%93a%3D-1%E6%97%B6%2Ca%26%23178%3B-4%3D-3%E6%AD%A4%E6%97%B6%E9%9B%86%E5%90%88%E4%B8%BA%7B-4%2C-3%2C-3%7D%E4%B8%8D%E6%BB%A1%E8%B6%B3%E9%9B%86%E5%90%88%E4%B8%AD%E5%85%83%E7%B4%A0%E7%9A%84%E4%BA%92%E5%BC%82)
若-3∈{a-3,2a-1,a²-4},求实数a的值.当a-3=-3时a=0当a=0时,2a-1=-1当a=0时,a²-4=-4此时集合为{-3,-1,-4},符合当2a-1=-3时,a=-1当a=-1时,a-3=-4当a=-1时,a²-4=-3此时集合为{-4,-3,-3}不满足集合中元素的互异
若-3∈{a-3,2a-1,a²-4},求实数a的值.
当a-3=-3时a=0
当a=0时,2a-1=-1
当a=0时,a²-4=-4
此时集合为{-3,-1,-4},符合
当2a-1=-3时,a=-1
当a=-1时,a-3=-4
当a=-1时,a²-4=-3
此时集合为{-4,-3,-3}
不满足集合中元素的互异性,舍去
当a²-4=-3时,
a=±1
为什么在代入的时候只将1里面代(为什么-1被舍去了)?
若-3∈{a-3,2a-1,a²-4},求实数a的值.当a-3=-3时a=0当a=0时,2a-1=-1当a=0时,a²-4=-4此时集合为{-3,-1,-4},符合当2a-1=-3时,a=-1当a=-1时,a-3=-4当a=-1时,a²-4=-3此时集合为{-4,-3,-3}不满足集合中元素的互异
因为a=-1在第二种情况(当2a-1=-3时,a=-1)中已经讨论过了.
a=-1时,2a-1=a²-4,需要舍去
a=-1时,在第二步已经讨论过了,无需再第三步中加以重复。
-1在第二种情况中出现过了